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I will keep it short and take only an extract (most important part) of the old task.

$$\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$$

What I have done is a lot work and time consuming, I have "simply" solved it. But I think with a lot less work, there would be an easier and faster way. It's just I cannot see it : /

If anyone wants see, here is my long solution which I'm not happy with:

$$\frac{n(n+1)(2n+1)+6(n+1)^2}{6}=\frac{(n^2+2n+n+2)(2n+3)}{6} \Leftrightarrow$$

$$\Leftrightarrow \frac{(2n^3+n^2+2n^2+n)+6n^2+12n+6}{6} = \frac{(n^2+3n+2)(2n+3)}{6} \Leftrightarrow$$

$$\Leftrightarrow \frac{2n^3+3n^2+n+6n^2+12n+6}{6}=\frac{2n^3+3n^2+6n^2+9n+4n+6}{6} \Leftrightarrow$$

$$\Leftrightarrow \frac{2n^3+9n^2+13n+6}{6}=\frac{2n^3+9n^2+13n+6}{6}$$

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    $\begingroup$ Factor out the common factor $(n+1)$ first and then further simplify and factorize. $\endgroup$ Feb 9, 2017 at 22:03
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    $\begingroup$ Verify that the equality holds for for distinct values of $n$, e.g., for $n=-2$, $n=-1$, $n=0$, $n=1$. As the difference between left and right is certainly a polynomial in $n$ of degree at most three and has four roots, it must be identically zero $\endgroup$ Feb 9, 2017 at 22:07
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    $\begingroup$ This is not a good what to organize a proof of equality: It should say $$ \frac{n(n+1)(2n+1)}{6}+(n+1)^2= \cdots\cdots\cdots = \frac{(n+1)(n+2)(2n+3)}{6} $$ with the blanks appropriately filled in. $\endgroup$ Feb 9, 2017 at 22:11
  • $\begingroup$ @MichaelHardy Does that mean I should better replace all the equivalence signs with equal signs? $\endgroup$
    – cnmesr
    Feb 9, 2017 at 22:14
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    $\begingroup$ Yes, but not in the same places where the equivalence signs appear. Put "equals" signs only between things you've already shown to be equal, not between things you're trying to prove to be equal. $\endgroup$ Feb 9, 2017 at 22:20

5 Answers 5

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Since you want $n+1$ to appear as a factor at the end, I would just leave that as a factor: $$ \frac{n(n+1)(2n+1)+6(n+1)^2}6 = \frac {n(2n+1) + 6(n+1)} 6 (n+1). $$ That becomes $$ \frac {2n^2 + n + 6n + 6} 6 (n+1) = \frac{2n^2 + 7n+6} 6 (n+1) = \frac{(n+2)(2n+3)} 6 (n+1) $$

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Oh goodness, you needn't so much work. All you need to do is this:

$$\begin{align}\frac{n(n+1)(2n+1)}6+(n+1)^2&=\frac{n(n+1)(2n+1)}6+\frac{6(n+1)^2}6\\&\tag 1=c\bigg(n(2n+1)+6(n+1)\bigg)\\&\tag2=c\bigg(2n^2+7n+6\bigg)\\&\tag3=c\bigg((n+2)(2n+3)\bigg)\end{align}$$

$(1)$ factor out $(n+1)/6$ and call it $c$.

$(2)$ expand the remaining stuff

$(3)$ factor the remaining stuff.

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Here is one solution. Since each side of identity is a polynomial of third degree (you can reduce it to degree 2 by canceling $(n+1)$), you can just verify that it holds for $4$ (or just $3$ after cancellation) distinct values of $n$ say $n=1,0,-1,-2$ then you are done. Note that I have chosen specific values of $n$ in such a manner to make calculations easy (some terms vanish for $n=0,-1,-2$).

This is useful if expressions are complicated. See the end of this answer https://math.stackexchange.com/a/1814894/72031 where I use this technique to prove an identity involving polynomials of degree $4$.

Update: this is identical to Hagen von Eitzen's comment which I saw later. I am marking it as community wiki.

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  • $\begingroup$ I think this is no reason to mark this as community wiki. If Hagen von Eitzen does not write an answer but you do, than it is ok. Actually it is not a good idea to answer a question in a comment as Hagen does. $\endgroup$
    – miracle173
    Feb 10, 2017 at 8:33
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Note that $$n(2n+1)+6(n+1)=2n^2+7n+6=(n+2)(2n+3)$$ Multiplying both sides by $\dfrac {n+1}6$ $$\frac {n(n+1)(2n+1)}6+(n+1)^2=\frac {(n+1)(n+2)(2n+3)}6$$

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The most general method: do all multiplications on both sides to transform both numerators to canonical form, then verify polynomials are the same degree and have respective coefficients equal.

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